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The Four Survivalists
The Four Survivalists, also known as the List Problem, is a math problem devised by Gilbert Martinez in October 2015. It is still being worked on by Martinez. The Problem 'In Numbers' How many ways can you write a list containing the numbers 1, 2, 3, and 4 such that the list satisfies the following conditions: *At no point does the same number appear consecutively in the list. *All numbers are shown in the list the same number of times. Moreover, how many ways are there to write such a list overall? 'In Words' The word version of this problem gives rise to the problem's name. Zeke, Will, Frank, and Carol are away on a survival trek. Each of them are assigned a certain task every day. The tasks are: cook, scout, guard, and hunter. Here are the following conditions: # A period (p'') is the number of days that pass before each person is assigned a task a certain number (''n) of times. # No one can hold the same position for two consecutive days until a period is concluded. Here are the questions: # How many possible configurations can be developed to assign each person one task, given they can be assigned a task n'' times? # How many configurations follow the conditions listed above? 'Definitions' Based on the conditions, here's how Martinez defined a period and a configuration. 'Period' : p = 4n Let ''n be the number of times a person is assigned a task. 'Configuration' : C_{n} = \frac{p!}{n}\, = \frac{(4n)!}{n}\, Calculations 'Let ''n = 1' : p = 4n = 4*1 = 4 : C_{1} = \frac{4!}{1}\, = 4! = 24 Here are all of the configurations for ''n = 1, which we will call c''1. ZWFC, ZCFW, ZFCW, ZWCF, ZCWF, ZFWC, WZFC, WCFZ, WFCZ, WZCF, WCZF, WFZC, FZCW, GWCZ, FCWZ, FZWC, FWZC, FCZW, CZFW, CWFZ, CFWZ, CZWF, CWZF, and CFZW. Thus, |''c''1| = 24. All 24 configurations follow the conditions. 'Let n'' = 2' : p = 4n = 4*2 = 8 : C_{2} = \frac{8!}{2}\, = \frac{40,320}{2}\, = 20,160 Using c''1, there are 408 configuration permutations that follow the conditions. For every configuration in ''c''1, there are 17 ways to create a unique permutation. There are 24 configurations in ''c''1, therefore 17 * 24 = 408. There are 24 additional configurations that can be created with the following 12 items: ZWZW, ZFZF, ZCZC, WZWZ, WFWF, WCWC, FZFZ, FWFW, FCFC, CZCZ, CWCW, and CFCF. For each of these items, there are 2 other items that can be affixed to it, creating 2 unique permutations. There are 12 items, therefore 12 * 2 = 24. Adding these permutations together results in 432 configurations, each of which will be grouped into ''c''2; |''c''2| = 432. 'Let n'' = 3' : p = 4n = 4*3 = 12 : C_{3} = \frac{12!}{3}\, = \frac{479,001,600}{3}\, = 159,667,200 As of now, |''c''3| is unknown. 'Lower Estimate' The lower estimate of the value of c''3 is found with the rate of change between the values of ''C''2 and ''C''3. : \Delta(C_{2}-C_{3}) = \frac{C_{3}-C_{2}}{C_{2}}\, = \frac{159,647,040}{20,160}\, = 7,919 The rate of change is then multiplied by the value of ''c''2 (432), giving it a proportional change equivalent to the change between ''C''2 and ''C''3. This gives us an estimate of the value of ''c''3. : |c_{3}| \approx \Delta(C_{2}-C_{3})c_{2} = 7,919*432 = 3,421,008 The lower estimate of |''c''3| is 3,421,008. 'Upper Estimate' The upper estimate of |''c''3| is found by the quotient of ''c''2 and ''C''2. : \frac{c_{2}}{C_{2}}\, = \frac{432}{20,160}\, = \frac{3}{140}\, The quotient is then multiplied by the value of ''C''3. : |c_{3}| \approx \frac{3C_{3}}{140}\, = 3,421,440 The upper estimate of |''c''3| is 3,421,440. 3,421,008 ≤ |''c''3| ≤ 3,421,440. A more precise estimate of |''c''3| can be found by the average of these two numbers. : |c_{3}| \approx \frac{3,421,008 + 3,421,440}{2}\, = 3,421,224 Therefore, |''c''3| ≈ 3,421,224. 'Let n'' = 4' : p = 4n = 4*4 = 16 : C_{4} = \frac{16!}{4}\, = \frac{20,922,789,888,000}{4}\, = 5,230,697,472,000 As of now, the value of |''c''4| is unknown. 'Lower Estimate' The lower estimate of the value of c''4 is found with the rate of change between the values of ''C''3 and ''C''4. : \Delta(C_{3}-C_{4}) = \frac{C_{4}-C_{3}}{C_{3}}\, = \frac{5,230,537,804,800}{159,667,200}\, = 32,759 The rate of change is then multiplied by the lower estimate of value of ''c''3 (3,421,440), giving it a proportional change equivalent to the change between ''C''3 and ''C''4. This gives us an estimate of |''c''4|. : |c_{4}| \approx \Delta(C_{3}-C_{4})c_{3} = 32,759*3,421,008 = 112,068,801,072 The lower estimate of the value of ''c''3 is 112,068,801,072. 'Upper Estimate' The upper estimate of the value ''c''4 is found by the quotient of ''c''2 and ''C''2. The quotient is then multiplied by the value of ''C''4. : |c_{4}| \approx \frac{3C_{4}}{140}\, = 112,086,374,400 The upper estimate of |''c''4| is 112,086,374,400. 112,068,801,072 ≤ |''c''4| ≤ 112,086,374,400. A more precise estimate of |''c''4| can be found by the average of these two numbers. : |c_{4}| \approx \frac{112,068,801,072 + 112,086,374,400}{2}\, = 112,077,587,736 Therefore, |''c''4| ≈ 112,077,587,736. General Formulas For the total configurations, the general formula is: : C_{n} = \frac{(xn)!}{n}\, Here, ''x is the number of unique items in the list, and n'' is the number of times each item must appear in the list. As of now, there is no general formula for the exact value of |''cn''|. However, there is a formula for an estimation of the value of |''cn''|: : |c_{n}| \approx \frac{\Delta(C_{n-1}-C_{n})c_{n-1}+\frac{3C_{n}}{140}\,}{2}\, This formula was devised by Martinez on April 30, 2016. Of course, this formula assumes ''x = 4. Additional Questions With this question, Martinez also asks if the values for each of these configurations can be described by a specific function; he also asks what the proportions are between the conditioned configurations and the possible configurations as n gets bigger. Martinez is currently working on finding the answers to these questions.